A curve has equation $y = x^5 + 4x^3 + 7x + q$ where $q$ is a positive constant - AQA - A-Level Maths Mechanics - Question 2 - 2018 - Paper 3

To find the gradient of the curve, we first need to calculate the derivative of the function with respect to $x$.

The function is given by:

$y = x^5 + 4x^3 + 7x + q$

Taking the derivative, we get:

$\frac{dy}{dx} = 5x^4 + 12x^2 + 7$

Next, we substitute $x = 0$ into the derivative to find the gradient at that point:

$\frac{dy}{dx}\bigg|_{x=0} = 5(0)^4 + 12(0)^2 + 7 = 7$

Thus, the gradient of the curve at the point where $x = 0$ is $7$. Therefore, we circle the answer: 7.